A point Mass inside a spherical shell



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We assumed at the beginning that the point mass m was outside the spherical shell, so our proof is valid only when m is outside a spherically symmetric mass distribution. When m is inside a spherical shell, the geometry is as shown in 1 Fig. 13.24.

 

 

 

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1 Kepler's first law link
2 Kepler’s second Law link
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3 Kepler’s third Law link
4 Planetary Motions and the Center of Mass link
5 Spherical Mass Distributions link

 

 

 

 

 

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1 GASES LIQUID AND DENSITY link
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3 pressure, depth, and pascals Law link
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The entire analysis goes just as before; Eqs. (13.18) through (13.22) are still valid. But when we get to Eq. (13.23), the limits of integration have to be changed to R - r and R + r. We then have

These Topics Are Also In Your Syllabus
1 Pressure in a fLuid link
2 pressure, depth, and pascals Law link
You May Find Something Very Interesting Here. link
3 PASCAL LAW link
4 absolute Pressure and Gauge Pressure link
5 Types Of Systems link

 

 

 

and the final result is

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Compare this result to Eq. (13.24): Instead of having r, the distance between m and the center of M, in the denominator, we have R, the radius of the shell. This means that U in Eq. (13.26) doesn’t depend on r and thus has the same value everywhere inside the shell. When m moves around inside the shell, no work is done on it, so the force on m at any point inside the shell must be zero.

More generally, at any point in the interior of any spherically symmetric mass distribution (not necessarily a shell), at a distance r from its center, the gravitational force on a point mass m is the same as though we removed all the mass at points farther than r from the center and concentrated all the remaining mass at the center.


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Rating - 4/5